Core Pure · Argand Diagrams · Year 1

Modulus-Argument Form

Rewriting a complex number as a modulus and an angle turns multiplication and division into simple addition and subtraction of angles.

16 min read 4 worked examples

Writing complex numbers in modulus-argument form

Definition 5.1 — modulus-argument (polar) form

Any complex number $z=a+bi$ with modulus $r=|z|$ and argument $\theta=\arg(z)$ can be written $$z=r(\cos\theta+i\sin\theta)$$ since $a=r\cos\theta$ and $b=r\sin\theta$. This is often abbreviated $z=r\operatorname{cis}\theta$.

An Argand diagram showing a complex number as a point with a line from the origin labelled r, and the angle theta marked between this line and the positive real axis, with dashed lines showing the real and imaginary components

Fig. 1 — the modulus r and argument θ locate a point exactly as well as its Cartesian coordinates a = r cos θ, b = r sin θ.

Example 1 · Converting to modulus-argument form

Write $z=-\sqrt3+i$ in the form $r(\cos\theta+i\sin\theta)$, using an exact angle.

$r=\sqrt{3+1}=2$. Reference angle: $\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=\dfrac{\pi}{6}$. Since $a<0,b>0$ (second quadrant), $\theta=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$.

$$z=2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)$$

Example 2 · Converting back to Cartesian form

Write $z=6\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)$ in the form $a+bi$, using exact values.

$\cos\dfrac{2\pi}{3}=-\dfrac12$, $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$.

$z=6\left(-\dfrac12+i\dfrac{\sqrt3}{2}\right)=-3+3\sqrt3\,i$.

Multiplying and dividing in modulus-argument form

Theorem 5.2 — multiplication and division rules

If $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$, then $$z_1z_2=r_1r_2\big(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big),$$ $$\frac{z_1}{z_2}=\frac{r_1}{r_2}\big(\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\big)$$ In words: multiply moduli, add arguments; divide moduli, subtract arguments.

ProofThe multiplication rule

$z_1z_2=r_1r_2(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)$

Substitute the modulus-argument forms.

$=r_1r_2\big[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2)\big]$

Expand the brackets and collect real/imaginary parts using $i^2=-1$.

$=r_1r_2\big[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big]$

The compound angle identities $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\sin(A+B)=\sin A\cos B+\cos A\sin B$.

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Example 3 · Multiplying in modulus-argument form

Given $z_1=4\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$ and $z_2=3\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$, find $z_1z_2$ in modulus-argument form.

$|z_1z_2|=4\times3=12$. $\arg(z_1z_2)=\dfrac{\pi}{4}+\dfrac{\pi}{6}=\dfrac{3\pi}{12}+\dfrac{2\pi}{12}=\dfrac{5\pi}{12}$.

$$z_1z_2=12\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)$$

Example 4 · Dividing in modulus-argument form

Using the same $z_1,z_2$ as Example 3, find $\dfrac{z_1}{z_2}$ in modulus-argument form.

$\left|\dfrac{z_1}{z_2}\right|=\dfrac43$. $\arg\left(\dfrac{z_1}{z_2}\right)=\dfrac{\pi}{4}-\dfrac{\pi}{6}=\dfrac{3\pi}{12}-\dfrac{2\pi}{12}=\dfrac{\pi}{12}$.

$$\frac{z_1}{z_2}=\frac43\left(\cos\frac{\pi}{12}+i\sin\frac{\pi}{12}\right)$$

ProofThe division rule

$\dfrac{z_1}{z_2}=\dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2(\cos\theta_2+i\sin\theta_2)}\times\dfrac{\cos\theta_2-i\sin\theta_2}{\cos\theta_2-i\sin\theta_2}$

Multiply top and bottom by the conjugate of the denominator's trig factor, so the denominator becomes real.

Denominator: $r_2(\cos^2\theta_2+\sin^2\theta_2)=r_2$

Using $\cos^2\theta_2+\sin^2\theta_2\equiv1$.

Numerator: $r_1\big[(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2)\big]$

Expand the product and simplify using $i^2=-1$.

$\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\big[\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\big]$

The compound angle identities $\cos(A-B)=\cos A\cos B+\sin A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$.

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Common mistake

The multiplication/division rules apply only when both numbers are already in modulus-argument form with the same trig functions $\cos\theta+i\sin\theta$ — you cannot add arguments of numbers still written in Cartesian form without converting first.

Watch out — arguments not already in matching form

If a number is given as $r(\cos\theta-i\sin\theta)$ (a minus sign) it is not yet in standard modulus-argument form. Rewrite it first using $\cos\theta=\cos(-\theta)$ and $-\sin\theta=\sin(-\theta)$, giving $r(\cos(-\theta)+i\sin(-\theta))$, before applying the multiplication or division rules.

Example 5 · Rewriting before multiplying

Given $z_1=5\left(\cos\dfrac{\pi}{9}+i\sin\dfrac{\pi}{9}\right)$ and $z_2=2\left(\cos\dfrac{\pi}{5}-i\sin\dfrac{\pi}{5}\right)$, find $z_1z_2$ in the form $x+iy$.

$z_2$ has a minus sign, so it isn't in standard form yet. Since $\cos\theta=\cos(-\theta)$ and $-\sin\theta=\sin(-\theta)$: $z_2=2\left(\cos\left(-\dfrac{\pi}{5}\right)+i\sin\left(-\dfrac{\pi}{5}\right)\right)$.

Now both are in standard form: modulus $=5\times2=10$; argument $=\dfrac{\pi}{9}-\dfrac{\pi}{5}=\dfrac{5\pi}{45}-\dfrac{9\pi}{45}=-\dfrac{4\pi}{45}$.

$z_1z_2=10\left(\cos\left(-\dfrac{4\pi}{45}\right)+i\sin\left(-\dfrac{4\pi}{45}\right)\right)\approx9.61-3.90i$ (to 3 s.f.).

Exam-style questions

Question 1 5 marks

Write $z=4-4i$ in modulus-argument form, using an exact angle.

$r=\sqrt{16+16}=4\sqrt2$ (M1, A1)

Fourth quadrant, reference angle $\tan^{-1}(1)=\pi/4$, so $\theta=-\pi/4$ (M1, A1)

$z=4\sqrt2\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right)$ (A1)

Question 2 6 marks

Given $z_1=8\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$ and $z_2=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$, find $z_1z_2$ and $\dfrac{z_1}{z_2}$ in modulus-argument form, using principal arguments.

$z_1z_2$: modulus $=16$; argument $=\dfrac{5\pi}{6}+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+\dfrac{2\pi}{6}=\dfrac{7\pi}{6}$, which is outside $(-\pi,\pi]$, so subtract $2\pi$: $\dfrac{7\pi}{6}-2\pi=-\dfrac{5\pi}{6}$ (M1, A1, M1, A1)

$z_1z_2=16\left(\cos\left(-\dfrac{5\pi}{6}\right)+i\sin\left(-\dfrac{5\pi}{6}\right)\right)$

$\dfrac{z_1}{z_2}$: modulus $=4$; argument $=\dfrac{5\pi}{6}-\dfrac{\pi}{3}=\dfrac{\pi}{2}$ (A1, A1)

$\dfrac{z_1}{z_2}=4\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)$

Question 3 5 marks

Given $z=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$, find $z^2$ in modulus-argument form, and hence in the form $a+bi$.

$z^2=z\times z$: modulus $=2\times2=4$; argument $=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$ (M1, A1)

$z^2=4\left(\cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)=4\left(-\dfrac12+i\dfrac{\sqrt3}{2}\right)=-2+2\sqrt3\,i$ (M1, A1, A1)

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